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Question: How do you solve and find the value of \(\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \rig...

How do you solve and find the value of sin⁡(2sin⁡−1(12))\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) ?

Explanation

Solution

We are asked to solve and find the value of sin⁡(2sin⁡−1(12))\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) , we start our solution by considering sin⁡−1(12){{\sin }^{-1}}\left( \dfrac{1}{2} \right) as θ\theta , then of the that we use sin⁡(2θ)=2sin⁡θcos⁡θ\sin \left( 2\theta \right)=2\sin \theta \cos \theta identity, after that we will learn how the sin and sin⁡−1{{\sin }^{-1}} are connected to each other, which we use further to solve the problem, we will also use sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 to change the θ\theta to cos theta or sin theta into one another.

Complete step by step solution:
We are given a function as sin⁡(2sin⁡−1(12))\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) , we have to evaluate the value of this function. To do so, we will learn that trigonometric functions are related to its inverses and we can also convert one trigonometric function to another.
Now as we have sin⁡(2sin⁡−1(12))\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) .
So, we consider sin⁡−1(12){{\sin }^{-1}}\left( \dfrac{1}{2} \right) as θ\theta then our equation become –
sin⁡(2sin⁡−1(12))=sin⁡(2θ)\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=\sin \left( 2\theta \right)
Now, we know that –
sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta
So, we get –
sin⁡(2sin⁡−1(12))=2sin⁡(sin⁡−1(12)).cos⁡(sin⁡−1(12))\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=2\sin \left( {{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right).\cos \left( {{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) …………………………………. (1)
Now we learn about how the ratio and its inverse are connected to each other.
For sin we have that sin⁡(sin⁡−1θ)=θ\sin \left( {{\sin }^{-1}}\theta \right)=\theta
And similarly for cos, we have cos⁡(cos⁡−1θ)=θ\cos \left( {{\cos }^{-1}}\theta \right)=\theta in equation (1), we have sin⁡(sin⁡−1(12))\sin \left( {{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) .
So, using above identity we get –
sin⁡(2sin⁡−1(12))=12\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=\dfrac{1}{2}
For cos⁡(2sin⁡−1(12))\cos \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right) we convert sin⁡−1(12){{\sin }^{-1}}\left( \dfrac{1}{2} \right) into cos⁡−1{{\cos }^{-1}} .
Now, as we know, θ=sin⁡−112\theta ={{\sin }^{-1}}\dfrac{1}{2} so, sin⁡θ=12\sin \theta =\dfrac{1}{2}
We know cos⁡2θ+sin⁡2θ=1{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1
Hence cos⁡θ=±1−sin⁡2θ\cos \theta =\pm \sqrt{1-{{\sin }^{2}}\theta } .
Hence, we get –
cos⁡θ=±1−sin⁡2θ\cos \theta =\pm \sqrt{1-{{\sin }^{2}}\theta }
As sin⁡2θ=12{{\sin }^{2}}\theta =\dfrac{1}{2} so,
cos⁡θ=±1−(12)2\cos \theta =\pm \sqrt{1-{{\left( \dfrac{1}{2} \right)}^{2}}}
By simplifying, we get –
cos⁡θ=±34\cos \theta =\pm \sqrt{\dfrac{3}{4}} (as 1−(12)2=1−14=341-{{\left( \dfrac{1}{2} \right)}^{2}}=1-\dfrac{1}{4}=\dfrac{3}{4} )
So, cos⁡θ=±32\cos \theta =\pm \dfrac{\sqrt{3}}{2}
Hence, θ=cos⁡−1(±32)\theta ={{\cos }^{-1}}\left( \pm \dfrac{\sqrt{3}}{2} \right)
So, we get –
sin⁡−1(12)=cos⁡−1(±32){{\sin }^{-1}}\left( \dfrac{1}{2} \right)={{\cos }^{-1}}\left( \pm \dfrac{\sqrt{3}}{2} \right)
Thus cos⁡(sin⁡−1(12))=cos⁡(cos⁡−1(±32))\cos \left( {{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=\cos \left( {{\cos }^{-1}}\left( \pm \dfrac{\sqrt{3}}{2} \right) \right)
We get –
=±32=\pm \dfrac{\sqrt{3}}{2}
So, we get two θ\theta values
Case (I) When cos⁡.sin⁡−1(12)=+32\cos .{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=+\dfrac{\sqrt{3}}{2}
Then as sin⁡−1(sin⁡(12))=12{{\sin }^{-1}}\left( \sin \left( \dfrac{1}{2} \right) \right)=\dfrac{1}{2} and cos⁡(sin⁡−1(12))=32\cos \left( {{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=\dfrac{\sqrt{3}}{2}
So using these in equation (1) we get –
sin⁡(2sin⁡−1(12))=2×12×32\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=2\times \dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}
=32=\dfrac{\sqrt{3}}{2}
Case (II) when cos⁡(sin⁡−1(12))=−32\cos \left( {{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=-\dfrac{\sqrt{3}}{2} as sin⁡−1(sin⁡(12))=12{{\sin }^{-1}}\left( \sin \left( \dfrac{1}{2} \right) \right)=\dfrac{1}{2} and cos⁡(sin⁡−1(12))=−32\cos \left( {{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=-\dfrac{\sqrt{3}}{2}
We get by using these in equation (1)
sin⁡(2sin⁡−1(12))=2×12×−32\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=2\times \dfrac{1}{2}\times \dfrac{-\sqrt{3}}{2}
=−32=-\dfrac{\sqrt{3}}{2}
Hence, we get –
sin⁡(2sin⁡−1(12))=±32\sin \left( 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right)=\pm \dfrac{\sqrt{3}}{2}.

Note: While solving such problems we should be very careful with identity like sin⁡2x≠2sin⁡x\sin 2x\ne 2\sin x or cos⁡2θ≠2cos⁡θsin⁡θ\cos 2\theta \ne 2\cos \theta \sin \theta .
We should not mix or use appropriate identity; also we should always cross check solutions so that change at error will get eliminated.
While solving fractions we always report answers in the simplest form so if something is common we need to cancel it always.